\(\int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx\) [488]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=-\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {2 (8 A b-11 a B) (a+b x)^{3/2}}{99 a^2 x^{9/2}}-\frac {4 b (8 A b-11 a B) (a+b x)^{3/2}}{231 a^3 x^{7/2}}+\frac {16 b^2 (8 A b-11 a B) (a+b x)^{3/2}}{1155 a^4 x^{5/2}}-\frac {32 b^3 (8 A b-11 a B) (a+b x)^{3/2}}{3465 a^5 x^{3/2}} \]

[Out]

-2/11*A*(b*x+a)^(3/2)/a/x^(11/2)+2/99*(8*A*b-11*B*a)*(b*x+a)^(3/2)/a^2/x^(9/2)-4/231*b*(8*A*b-11*B*a)*(b*x+a)^
(3/2)/a^3/x^(7/2)+16/1155*b^2*(8*A*b-11*B*a)*(b*x+a)^(3/2)/a^4/x^(5/2)-32/3465*b^3*(8*A*b-11*B*a)*(b*x+a)^(3/2
)/a^5/x^(3/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {79, 47, 37} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=-\frac {32 b^3 (a+b x)^{3/2} (8 A b-11 a B)}{3465 a^5 x^{3/2}}+\frac {16 b^2 (a+b x)^{3/2} (8 A b-11 a B)}{1155 a^4 x^{5/2}}-\frac {4 b (a+b x)^{3/2} (8 A b-11 a B)}{231 a^3 x^{7/2}}+\frac {2 (a+b x)^{3/2} (8 A b-11 a B)}{99 a^2 x^{9/2}}-\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}} \]

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^(13/2),x]

[Out]

(-2*A*(a + b*x)^(3/2))/(11*a*x^(11/2)) + (2*(8*A*b - 11*a*B)*(a + b*x)^(3/2))/(99*a^2*x^(9/2)) - (4*b*(8*A*b -
11*a*B)*(a + b*x)^(3/2))/(231*a^3*x^(7/2)) + (16*b^2*(8*A*b - 11*a*B)*(a + b*x)^(3/2))/(1155*a^4*x^(5/2)) - (
32*b^3*(8*A*b - 11*a*B)*(a + b*x)^(3/2))/(3465*a^5*x^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {\left (2 \left (-4 A b+\frac {11 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx}{11 a} \\ & = -\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {2 (8 A b-11 a B) (a+b x)^{3/2}}{99 a^2 x^{9/2}}+\frac {(2 b (8 A b-11 a B)) \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx}{33 a^2} \\ & = -\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {2 (8 A b-11 a B) (a+b x)^{3/2}}{99 a^2 x^{9/2}}-\frac {4 b (8 A b-11 a B) (a+b x)^{3/2}}{231 a^3 x^{7/2}}-\frac {\left (8 b^2 (8 A b-11 a B)\right ) \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx}{231 a^3} \\ & = -\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {2 (8 A b-11 a B) (a+b x)^{3/2}}{99 a^2 x^{9/2}}-\frac {4 b (8 A b-11 a B) (a+b x)^{3/2}}{231 a^3 x^{7/2}}+\frac {16 b^2 (8 A b-11 a B) (a+b x)^{3/2}}{1155 a^4 x^{5/2}}+\frac {\left (16 b^3 (8 A b-11 a B)\right ) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{1155 a^4} \\ & = -\frac {2 A (a+b x)^{3/2}}{11 a x^{11/2}}+\frac {2 (8 A b-11 a B) (a+b x)^{3/2}}{99 a^2 x^{9/2}}-\frac {4 b (8 A b-11 a B) (a+b x)^{3/2}}{231 a^3 x^{7/2}}+\frac {16 b^2 (8 A b-11 a B) (a+b x)^{3/2}}{1155 a^4 x^{5/2}}-\frac {32 b^3 (8 A b-11 a B) (a+b x)^{3/2}}{3465 a^5 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=-\frac {2 (a+b x)^{3/2} \left (128 A b^4 x^4+35 a^4 (9 A+11 B x)+24 a^2 b^2 x^2 (10 A+11 B x)-16 a b^3 x^3 (12 A+11 B x)-10 a^3 b x (28 A+33 B x)\right )}{3465 a^5 x^{11/2}} \]

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x))/x^(13/2),x]

[Out]

(-2*(a + b*x)^(3/2)*(128*A*b^4*x^4 + 35*a^4*(9*A + 11*B*x) + 24*a^2*b^2*x^2*(10*A + 11*B*x) - 16*a*b^3*x^3*(12
*A + 11*B*x) - 10*a^3*b*x*(28*A + 33*B*x)))/(3465*a^5*x^(11/2))

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.67

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (128 A \,b^{4} x^{4}-176 B a \,b^{3} x^{4}-192 A a \,b^{3} x^{3}+264 B \,a^{2} b^{2} x^{3}+240 A \,a^{2} b^{2} x^{2}-330 B \,a^{3} b \,x^{2}-280 A \,a^{3} b x +385 B \,a^{4} x +315 A \,a^{4}\right )}{3465 x^{\frac {11}{2}} a^{5}}\)	\(101\)
default	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (128 A \,b^{4} x^{4}-176 B a \,b^{3} x^{4}-192 A a \,b^{3} x^{3}+264 B \,a^{2} b^{2} x^{3}+240 A \,a^{2} b^{2} x^{2}-330 B \,a^{3} b \,x^{2}-280 A \,a^{3} b x +385 B \,a^{4} x +315 A \,a^{4}\right )}{3465 x^{\frac {11}{2}} a^{5}}\)	\(101\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (128 A \,b^{5} x^{5}-176 B a \,b^{4} x^{5}-64 a A \,b^{4} x^{4}+88 B \,a^{2} b^{3} x^{4}+48 a^{2} A \,b^{3} x^{3}-66 B \,a^{3} b^{2} x^{3}-40 a^{3} A \,b^{2} x^{2}+55 B \,a^{4} b \,x^{2}+35 a^{4} A b x +385 a^{5} B x +315 a^{5} A \right )}{3465 x^{\frac {11}{2}} a^{5}}\)	\(125\)

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^(13/2),x,method=_RETURNVERBOSE)

[Out]

-2/3465*(b*x+a)^(3/2)*(128*A*b^4*x^4-176*B*a*b^3*x^4-192*A*a*b^3*x^3+264*B*a^2*b^2*x^3+240*A*a^2*b^2*x^2-330*B
*a^3*b*x^2-280*A*a^3*b*x+385*B*a^4*x+315*A*a^4)/x^(11/2)/a^5

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=-\frac {2 \, {\left (315 \, A a^{5} - 16 \, {\left (11 \, B a b^{4} - 8 \, A b^{5}\right )} x^{5} + 8 \, {\left (11 \, B a^{2} b^{3} - 8 \, A a b^{4}\right )} x^{4} - 6 \, {\left (11 \, B a^{3} b^{2} - 8 \, A a^{2} b^{3}\right )} x^{3} + 5 \, {\left (11 \, B a^{4} b - 8 \, A a^{3} b^{2}\right )} x^{2} + 35 \, {\left (11 \, B a^{5} + A a^{4} b\right )} x\right )} \sqrt {b x + a}}{3465 \, a^{5} x^{\frac {11}{2}}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(13/2),x, algorithm="fricas")

[Out]

-2/3465*(315*A*a^5 - 16*(11*B*a*b^4 - 8*A*b^5)*x^5 + 8*(11*B*a^2*b^3 - 8*A*a*b^4)*x^4 - 6*(11*B*a^3*b^2 - 8*A*
a^2*b^3)*x^3 + 5*(11*B*a^4*b - 8*A*a^3*b^2)*x^2 + 35*(11*B*a^5 + A*a^4*b)*x)*sqrt(b*x + a)/(a^5*x^(11/2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1413 vs. \(2 (150) = 300\).

Time = 92.48 (sec) , antiderivative size = 1413, normalized size of antiderivative = 9.42 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=\text {Too large to display} \]

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**(13/2),x)

[Out]

-630*A*a**9*b**(33/2)*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7
+ 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 2590*A*a**8*b**(35/2)*x*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x
**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 3980*A*a
**7*b**(37/2)*x**2*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 1
3860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 2716*A*a**6*b**(39/2)*x**3*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x
**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 686*A*a*
*5*b**(41/2)*x**4*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13
860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 70*A*a**4*b**(43/2)*x**5*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5
+ 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 560*A*a**3*
b**(45/2)*x**6*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860
*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 1120*A*a**2*b**(47/2)*x**7*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5
+ 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 896*A*a*b**(
49/2)*x**8*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a**8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**
6*b**19*x**8 + 3465*a**5*b**20*x**9) - 256*A*b**(51/2)*x**9*sqrt(a/(b*x) + 1)/(3465*a**9*b**16*x**5 + 13860*a*
*8*b**17*x**6 + 20790*a**7*b**18*x**7 + 13860*a**6*b**19*x**8 + 3465*a**5*b**20*x**9) - 70*B*a**7*b**(19/2)*sq
rt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) - 220*B
*a**6*b**(21/2)*x*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4
*b**12*x**7) - 228*B*a**5*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**
5*b**11*x**6 + 315*a**4*b**12*x**7) - 80*B*a**4*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**
6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) + 10*B*a**3*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(315*a*
*7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) + 60*B*a**2*b**(29/2)*x**5*sqr
t(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) + 80*B*a
*b**(31/2)*x**6*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b
**12*x**7) + 32*B*b**(33/2)*x**7*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*
x**6 + 315*a**4*b**12*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=\frac {32 \, \sqrt {b x^{2} + a x} B b^{4}}{315 \, a^{4} x} - \frac {256 \, \sqrt {b x^{2} + a x} A b^{5}}{3465 \, a^{5} x} - \frac {16 \, \sqrt {b x^{2} + a x} B b^{3}}{315 \, a^{3} x^{2}} + \frac {128 \, \sqrt {b x^{2} + a x} A b^{4}}{3465 \, a^{4} x^{2}} + \frac {4 \, \sqrt {b x^{2} + a x} B b^{2}}{105 \, a^{2} x^{3}} - \frac {32 \, \sqrt {b x^{2} + a x} A b^{3}}{1155 \, a^{3} x^{3}} - \frac {2 \, \sqrt {b x^{2} + a x} B b}{63 \, a x^{4}} + \frac {16 \, \sqrt {b x^{2} + a x} A b^{2}}{693 \, a^{2} x^{4}} - \frac {2 \, \sqrt {b x^{2} + a x} B}{9 \, x^{5}} - \frac {2 \, \sqrt {b x^{2} + a x} A b}{99 \, a x^{5}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{11 \, x^{6}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(13/2),x, algorithm="maxima")

[Out]

32/315*sqrt(b*x^2 + a*x)*B*b^4/(a^4*x) - 256/3465*sqrt(b*x^2 + a*x)*A*b^5/(a^5*x) - 16/315*sqrt(b*x^2 + a*x)*B
*b^3/(a^3*x^2) + 128/3465*sqrt(b*x^2 + a*x)*A*b^4/(a^4*x^2) + 4/105*sqrt(b*x^2 + a*x)*B*b^2/(a^2*x^3) - 32/115
5*sqrt(b*x^2 + a*x)*A*b^3/(a^3*x^3) - 2/63*sqrt(b*x^2 + a*x)*B*b/(a*x^4) + 16/693*sqrt(b*x^2 + a*x)*A*b^2/(a^2
*x^4) - 2/9*sqrt(b*x^2 + a*x)*B/x^5 - 2/99*sqrt(b*x^2 + a*x)*A*b/(a*x^5) - 2/11*sqrt(b*x^2 + a*x)*A/x^6

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=\frac {2 \, {\left ({\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (11 \, B a b^{10} - 8 \, A b^{11}\right )} {\left (b x + a\right )}}{a^{5}} - \frac {11 \, {\left (11 \, B a^{2} b^{10} - 8 \, A a b^{11}\right )}}{a^{5}}\right )} + \frac {99 \, {\left (11 \, B a^{3} b^{10} - 8 \, A a^{2} b^{11}\right )}}{a^{5}}\right )} - \frac {231 \, {\left (11 \, B a^{4} b^{10} - 8 \, A a^{3} b^{11}\right )}}{a^{5}}\right )} {\left (b x + a\right )} + \frac {1155 \, {\left (B a^{5} b^{10} - A a^{4} b^{11}\right )}}{a^{5}}\right )} {\left (b x + a\right )}^{\frac {3}{2}} b}{3465 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^(13/2),x, algorithm="giac")

[Out]

2/3465*((2*(b*x + a)*(4*(b*x + a)*(2*(11*B*a*b^10 - 8*A*b^11)*(b*x + a)/a^5 - 11*(11*B*a^2*b^10 - 8*A*a*b^11)/
a^5) + 99*(11*B*a^3*b^10 - 8*A*a^2*b^11)/a^5) - 231*(11*B*a^4*b^10 - 8*A*a^3*b^11)/a^5)*(b*x + a) + 1155*(B*a^
5*b^10 - A*a^4*b^11)/a^5)*(b*x + a)^(3/2)*b/(((b*x + a)*b - a*b)^(11/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{13/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{11}+\frac {x\,\left (770\,B\,a^5+70\,A\,b\,a^4\right )}{3465\,a^5}+\frac {x^5\,\left (256\,A\,b^5-352\,B\,a\,b^4\right )}{3465\,a^5}+\frac {4\,b^2\,x^3\,\left (8\,A\,b-11\,B\,a\right )}{1155\,a^3}-\frac {16\,b^3\,x^4\,\left (8\,A\,b-11\,B\,a\right )}{3465\,a^4}-\frac {2\,b\,x^2\,\left (8\,A\,b-11\,B\,a\right )}{693\,a^2}\right )}{x^{11/2}} \]

[In]

int(((A + B*x)*(a + b*x)^(1/2))/x^(13/2),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/11 + (x*(770*B*a^5 + 70*A*a^4*b))/(3465*a^5) + (x^5*(256*A*b^5 - 352*B*a*b^4))/(3465*
a^5) + (4*b^2*x^3*(8*A*b - 11*B*a))/(1155*a^3) - (16*b^3*x^4*(8*A*b - 11*B*a))/(3465*a^4) - (2*b*x^2*(8*A*b -
11*B*a))/(693*a^2)))/x^(11/2)